# Chapter 6: Annual Cash Flow ```144
Homework
Solutions
for
Engineering
Newnan,
'--
Chapter
6: Annual
Economic
Eschenbach,
Cash
Analysis,
9th Edition
Lavelle
Flow
Analysis
1\ ~;),11~l
6 -rn"
1
f~:~ie'~J
,.,c,'
\$60
~':1
\$45
\$30
\$15
;;
c
c
c
:
,-
c
, ""
C = \$15 + \$15 (AlG, 10%,4)
= \$15 + \$15 (1.381) = ~35.72
6
!
..":::'_1.._L..
-' 2
'-:la,
i~::)I
B
B
B
B
B
j
"
,,~
\$100
\$100
B = [\$100 + \$100 (F/P, 15%,4)] (AlF, 15%,5)
= [\$100 + \$100 (1.749)] (0.1483) = ~40.77
6-3
I
I
\$60
\$45
E
E
.\$30
i
j
j
E
E
--.'
E = \$60 -\$15 (AlG, 12%,4)
= \$60 -\$15 (1.359) = \$39.62
I
;i'
!
~
r
",./,
Homework Solutions for Engineering Economic Analysis, 9th Edition
Newnan, Eschenbach, Lavelle
6-32
Around the Lake
\$75,000
\$3,000/ r
\$7,500/ r
\$1,500/ r
\$45,000
15 ears
First Cost
Maintenance
Annual Power Loss
Pro en Taxes
Salva e Value
Useful Life
Under the Lake
\$125,000
\$2,000/ r
\$2,500/ r
\$2,500/ r
\$25,000
15 ears
Around the Lake
EUAC
= \$75,000 (AlP, 7%,15) + \$12,000 -\$45,000 (AlF, 7%,15)
= \$75,000 (0.1098) + \$12,000 -\$45,000 (0.0398)
= \$18.444
Under the Lake
EUAC
= \$125,000 (AlP, 7%,15) + \$7,000 -\$25,000 (AlF, 7%,15)
= \$125,000 (0.1098) + \$7,000 -\$25,000 (0.0398)
= \$19.730
Go around the lake.
6-33
:'~c' ~Jji 1(..i1
Engineering Department
Estimate
\$30
:
\$27
~
i
'
If.;
.\$9
\$24
Amounts x 103
\$21
\$] 8
\$15
\$12
IOE
\$6
EUAC
= \$30,000 -\$3,000 (AlG, 8%, 10)
= \$30,000 -\$3,000 (3.871)
= \$18,387
;~:(r:r
t:!Yro-clean'soffer of \$15.000/vr is less costlv.
"
'i
~
.
157
-160
Homework Solutions for Engineering Economic Analysis, 9th Edition
Newnan, Eschenbach,
Alternative
EUAB
i
A
-EUAC
= \$845
Alternative
B
EUAB
-EUAC
To
I
Lavelle
maximize
-\$3,000
= \$1,400
(EUAB
(0.30672)
-\$5,000
-EUAC)
= -\$75.16
(0.30672)
choose
= -\$133.60
alternative
A. (less
neaative
value).
6-37
Machine
X
EUAC
= \$5,000
(AlP,
= \$5,000
(0.2505)
8%,
5)
=\$1,252
Machine
Y
EUAC
= (\$8,000
=\$1,106
Select
Machine
-\$2,000)
(AlP,
8%,12)
+ \$2,000
(0.08)
+ \$150
!
..
Y.
6-38
Annual
Cost
of Diesel
Annual
Cost
of Gasoline
EUACdiesel
Fuel
= (\$13,000
= [\$50,000km/(35
km/i)]
x \$0.48/1
= \$685.71
= [\$50,000km/(28
km/i)]
x \$0.51/1
= \$910.71
-\$2,000)
+ \$685.71
= \$11,000
(AlP,
fuel
(0.2886)
6%,
+ \$300
+ \$120
4) + \$2,000
repairs
+ \$500
(0.06)
insurance
+ \$1,485.71
= \$4,780.31
EUACgasoline
= (\$12,000
-\$3,000)
+ \$910.71
(AlP,
fuel
6%,
+ \$200
3) + \$3,000
repairs
+ \$500
.
(0.06)
insurance
= \$5,157.61
The
diesel
taxi
is more
economical.
6-39
Machine
A
EUAC
= \$1,000
+ Pi
~i~-
= \$1,000
+ \$10,000
(AlP,
10%,4)
-\$10,000
(AlF,
10%,4)
=\$1,000+\$1,000
,
=\$2,000
(
,c..'cc
7
,
--
162
Homework Solutions for EngineeringEconomic Analysis, 9thEdition
Newnan, Eschenbach, Lavelle
6-42
I
Seven year analysis period:
Alternative A
EUAB -EUAC
= \$55 -[\$100
+ \$100 (P/F, 10%,3)
+ \$100 (P/F, 10%,6)] (AlP, 10%,7)
= \$55 -[\$100 + \$100 (0.7513) + \$100 (0.5645)] (0.2054)
= +\$7.43
Alternative B
EUAB -EUAC
= \$61 -[\$150
+ \$150 (P/F, 10%,4)] (AlP, 10%,7)
": \$61 -[\$150
+ \$150 (0.683)] (0.2054)
=~~
.Q1!QQ'§~
.
Note: The analysis periOd is seven years, hence one cannot compare three
§:
years of A vs. four years of B, If one does, the problem is Constructed
so he will get the wrong answer.
6-4S
6-43
Us
EUACgas ": (P -S) (AlP, 1'10,n) + SL + Annual Costs
= (\$2,400 -\$300) (AlP, 10%,5) + \$300 (0.10) + \$1,200 + \$300
= \$2,100 (0.2638) + \$30 + \$1,500
= \$2,084
-..
He:
Np'
EUACelectr= (\$6,000 -\$600) (AlP, 10%, 10) + \$600 (0.10) + \$750 + \$50
= \$5,400 (0.1627) + \$60 + \$800
= \$1,739
..
c
NPli\
'§~l~!Jb~m~mQ-mQ!Qr
.~
6-44
.QhQQ
EUAC Comparison
EqUlv
Gravity Plan
Initial Investment: = \$2.8 million (AlP, 10%, 40)
= \$2.8 million (0.1023)
Annual Operation and maintenance
Annual Cost
EUAC/
, I
"' .= \$286,400
~\ '
= \$10,000
= \$296,400
E
UACa
~
.QbQ~
)
Homework Solutions for Engineering Economic Analysis, 9th Edition
163
Newnan, Eschenbach, Lavelle
Pumping Plan
Initial Investment: = \$1.4 million (AlP, 10%, 20)
= \$1.4 million (0.1023)
= \$200,000 (P/F, 10%, 10) (AlP, 10%, 40)
= \$200,000 (0.3855) (0.1023)
= \$7,890
Annual Operation and maintenance
= \$25,000
Power Cost:
= \$50,000 for 40 years
= \$50,000
Additional Power Cost in last 30 years:
= \$50,000 (F/A, 10%,30) (AlF, 10%,40)
= \$50,000 (164.494) (0.00226)
= \$18,590
Annual Cost
= \$244,680
Select the Pumoing Plan."
= \$143,200
.
6-45
Use 20 year analysis period.
I
Net Present Worth Approach
~
NPWMas. = -\$250 -(\$250 -\$10) [(P/F, 6%,4) + (P/F, 6%, 8) + (P/F, 6%, 12)
+ (P/F, 6%,16)] + \$10 (P/F, 6%, 20) -\$20 (PIA, 6%, 20)
= -\$250 -\$240 [0.7921 + 0.6274 + 0.4970 + 0.3936)
+ \$10 (0.3118) -\$20 (11.470)
=-\$1,031
NPWBRK= -\$1,000 -\$10 (PIA, 6%, 20) + \$100 (P/F, 6%, 20)
= -\$1,000 -\$10 (11.470) + \$100 (0.3118)
= -\$1,083
~hoose Masonite to save \$52 on Present Worth of Cost.
Equivalent Uniform Annual Cost Approach
~
C,c'
EUACMas.
= \$20 + \$250 (AlP, 6%, 4) -\$10 (AlF, 6%, 4)
= \$20
+ \$250
(0.2886)
-\$10
!
(0.2286)
..I
'"
= \$90
EUACBRK
= \$10 + \$1,000 (AlP, 6%, 20) -\$100 (AlF, 6%, 20) .;.,,"
= \$10 + \$1,000 (,0.872) -\$100 (0.0272)"
= \$94
""!
(1:1;,
;~.,)i
"...I!J
.QhQose
Masonate to save \$4 oer ~ear.
d
1-
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~
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